Question

How do you integrate `int x^3sqrt(16-x^2)` by trigonometric substitution?

Answer 1

`intx^3sqrt(16-x^2)dx=-16/3(16-x^2)^(3/2)+1/5(16-x^2)^(5/2)+c`

Explanation:

Let `x=4sint`, then `sqrt(16-x^2)=4cost` and `dx=4costdt`

and `intx^3sqrt(16-x^2)dx`

= `int64sin^3t*4cost*4costdt`

= `1024intsin^3tcos^2tdt`

= `1024int(1-cos^2t)cos^2tsintdt`

= `1024int(cos^2t-cos^4t)sintdt`

Let `cost=u` then `du=-sintdt` and our integral becomes

`-1024int(u^2-u^4)du`

= `-1024(u^3/3-u^5/5)`

i.e. `1024(-1/3cos^3t+1/5cos^5t)` (substituting `u=cost`)

and as `cost=1/4sqrt(16-x^2)`, our integral becomes

`1024(-1/3*1/64(16-x^2)^(3/2)+1/5*1/1024(16-x^2)^(5/2))+c`

= `-16/3(16-x^2)^(3/2)+1/5(16-x^2)^(5/2)+c`

Answer 2

Let,`sqrt(16-x^2)=t=>16-x^2=t^2=>xdx=-tdt`
`I=int(16-t^2)t(-t)dt=int(t^4-16t^2)dt=t^5/5-(16t^3)/3+c`
Substituting, `t=sqrt(16-x^2)`
`I=(sqrt(16-x^2))^5/5-(16(sqrt(16-x^2))^3)/3+c`

Explanation:

Here,

`I=intx^3sqrt(16-x^2)dx`

Let, `x=4sinu=>dx=4cosudu`

`and sinu=x/4=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16`

`=>cosu=sqrt(16-x^2)/4...to(A)`

`I=int64sin^3usqrt(16-16sin^2u)xx4cosudu`

`=256intsin^3uxx4cosuxxcosudu`

`I=1024intsin^3ucos^2udu`

`I=4^5int(1-cos^2u)cos^2uxxsinudu`

`=4^5[intcos^2usinudu-intcos^4usinudu]`

`=4^5[-int(cosu)^2(-sinu)du+int(cosu)^4(-sinu)du]`

`=4^5[-(cosu)^3/3+(cosu)^5/5]+c`

`=4^5[(cosu)^5/5-(cosu)^3/3]+c...to`from `(A)`

`=4^5[(sqrt(16-x^2))^5/(5xx4^5)-(sqrt(16-x^2))^3/(3xx4^3)]+c`

`=(sqrt(16-x^2))^5/5-4^2xx(sqrt(16-x^2))^3/3+c`

`I=(sqrt(16-x^2))^5/5-(16(sqrt(16-x^2))^3)/3+c`